Show that sum of the two closed sets in $\mathbb{R^2}$ is not always closed.

In $\mathbb{R^2}$, consider the sets $A =\{(x,y)\,\,:\,\, x>0,\,\,xy=1\}$ and $B =\{(x,y)\,\,:\,\, x>0,\,\,xy=-1\}$ which are closed, show that the sum A + B is not closed. $$ A+B:=\{a+b : a \in A, b \in B\} $$

To have better intuition, you can visualize the two sets as the following picture

enter image description here

  • 3
    ${}$Nice example! – Lord Shark the Unknown Oct 10 at 18:45
  • 1
    What's the question? – Yanko Oct 10 at 18:56
  • I revised the question. Show that the set $A+B$ is not closed. – Saeed Oct 10 at 19:03
  • What subset of the plane is $A+B$? – rogerl Oct 10 at 19:56
  • What do you mean? – Saeed Oct 10 at 23:11
up vote 2 down vote accepted

Consider the sequence $$(x_n)=\Big\{\Big(\frac{2}{n},0\Big)\Big\}_{1}^\infty=\Big\{\Big(\frac{1}{n},n\Big)+\Big(\frac{1}{n},-n\Big)\Big\}_{1}^\infty \in A+B$$ Here $$x_n \longrightarrow (0,0) \notin A+B$$ since the first coordinate of the element in $A+B$ is always positive .

So, $A+B$ is not closed!

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